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PastaTapestry
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1. A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.9 m/s2. A green car arrives at the position of the stop-light 4 s after the light had turned green. What is the slowest constant speed which the green car can maintain and still catch up to the blue car?
Answer in units of m/s
So, we're given that ##a = 0.9 m/(s^2)##, ##v_0 = 0##, and that displacement at t = 0 is 0 (all of this for the blue car, or ##C_B##)
For the green car(or ##C_G##), we're given that the car is holding a constant speed, so there must be no acceleration. If there's no acceleration, then the final and initial velocities must be equal for ##C_G##. Additionally, we're given that its displacement = 0 at t = 4 (since it started 4 seconds late).
2. We know that the velocity for ##C_B## is $$v=0.9t$$ We know that the displacement is equal to $$0.45t^2$$
For ##C_G##, we know that ##v## is a constant c, that acceleration = 0, and that the displacement is represented by the function ##vt - 4v## (I believe).
3. I figured that we were looking for the minimum speed that ##C_G## would have to possesses to cause the displacement of both ##C_G## and ##C_B## to be equal, because this would represent that ##C_G## had caught up to ##C_B##. However, I kept getting it down to things like $$0.45t^2 = vt - 4v$$ which you can't solve to my knowledge. So I tried a graphical method. I know that the displacement of ##C_B## is a parabola, and that the displacement of ##C_G## is a linear equation that passes through the point (4,0) [if we graph displacement as y and time as x). So we're looking for a tangent that passes through the point (4,0). However, we don't know the time or displacement this occurs at.
Sorry if this is all too verbose and such, first time posting here so I apologize!
Answer in units of m/s
So, we're given that ##a = 0.9 m/(s^2)##, ##v_0 = 0##, and that displacement at t = 0 is 0 (all of this for the blue car, or ##C_B##)
For the green car(or ##C_G##), we're given that the car is holding a constant speed, so there must be no acceleration. If there's no acceleration, then the final and initial velocities must be equal for ##C_G##. Additionally, we're given that its displacement = 0 at t = 4 (since it started 4 seconds late).
2. We know that the velocity for ##C_B## is $$v=0.9t$$ We know that the displacement is equal to $$0.45t^2$$
For ##C_G##, we know that ##v## is a constant c, that acceleration = 0, and that the displacement is represented by the function ##vt - 4v## (I believe).
3. I figured that we were looking for the minimum speed that ##C_G## would have to possesses to cause the displacement of both ##C_G## and ##C_B## to be equal, because this would represent that ##C_G## had caught up to ##C_B##. However, I kept getting it down to things like $$0.45t^2 = vt - 4v$$ which you can't solve to my knowledge. So I tried a graphical method. I know that the displacement of ##C_B## is a parabola, and that the displacement of ##C_G## is a linear equation that passes through the point (4,0) [if we graph displacement as y and time as x). So we're looking for a tangent that passes through the point (4,0). However, we don't know the time or displacement this occurs at.
Sorry if this is all too verbose and such, first time posting here so I apologize!
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